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3 years ago

Consider this absolute value function.

Mathematics

1 answer:

densk [106]3 years ago

6 0

Answer:

Step-by-step explanation:

f(x) = |x+3|=x+3 when x+3≥0 ⇒x≥-3

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PLS HELP, Rectangle A'B'C'D' is the image of rectangle ABCD under a dilation. Which calculation of the area of rectangle A'B'C'D

aev [14]

What grade of math is this? I’ll answer I just need to know cause a lot of chapters could be related to this

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3 years ago

Find the solution to the differential equation dz/dt = 7te^4z that passes through the origin.

andreev551 [17]

Answer:

$z=(\frac{-1}{4} )ln|1-14t^{2}|\\$

Step-by-step explanation:

from the equation, $\frac{dz}{dt}=7te^{4z}\\$ .

we can give different approach to the equation, but to make it simple and direct, let separate the equation by bringing all like terms to the same side i.e

$\frac{dz}{e^{4z}}=7tdt\\e^{-4z} dz=7tdt$ .

if we integrate both side,

$\int\limits^a_b{e^{-4z} } \,dz =\int\limits^a_b {7t} \,dt$

$-1/4e^{-4z} +c_{1}=7/2t^{2} +c_{2}\\-1/4e^{-4z}= 7/2t^{2} +c_{2}-c_{1}\\let c_{2}-c_{1}=c \\-1/4e^{-4z}= 7/2t^{2} +c$

since the equation passes through the origin, and at the origin z=0 and t=0

we substitute this values and solve for the constant c

$e^{-4*0}= 7/2*0 +c\\c=1$ .

If we substitute the value of c into the equation we arrive at

$(-1/4)e^{-4z}= (7/2)t^{2}+1\\ e^{-4z}=1-14t^{2} \\$

if we the the natural logarithm of both sides, we arrive at

$-4z=ln|1-14t^{2}|\\z=(\frac{-1}{4} )ln|1-14t^{2}|\\$

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3 years ago

What is the range of this function {(3,4)(5,8),(9,-4),(-2,1),(0,8)}

denis23 [38]

To solve this problem you must apply the proccedure shown below:
1. You have the following information given in the problem above:
<span> {(3,4)(5,8),(9,-4),(-2,1),(0,8)}
2. Therefore,by definition, you have that the range is the set of all the values of y given. Keeping this on mind, you have that the range is:
Range:{4,8,-4,1,8}
Therefore, as you can see, the answer is: </span>Range:{4,8,-4,1,8}.

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3 years ago

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nika2105 [10]

Answer:

THE SECOND ONE

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3 years ago

Z is the center of the circle. Suppose that QR = 26, TV = 13, SV = 2x + 1, and VZ = 8. Find x and UZ.

Annette [7]

X=20
26+13+1+8=58

X=2 so 58/2 is 29 so x is 29. And to find the vz you just plug it in 29 into the equation 2x+1 2(29)+1=59

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3 years ago